/*
	Name: High_precision Plus
	Copyright: 2017
	Author: Fropised
	Date: 17/09/17 18:41
	Description: To plus two large numbers
*/
#include<iostream> //the headfile to input and output
#include<string> //the headfile to use the modal string
#include<algorithm> //the headfile to use the functions like reverse
#include<cstring> //the headfile to use memset
#define MaxN 200 //the maximum length of the two numbers
using namespace std;

string add(string s1,string s2) //the function to do the work
{
	int a[MaxN + 1],b[MaxN + 1],c[MaxN + 1]; //the array a,b are used to record the two strings
	//the array c is used to record the answer
	memset(a,0,sizeof(a));
	memset(b,0,sizeof(b));
	memset(c,0,sizeof(c));
	//to fill the three arrays with zero
	for (int i = 0;i < s1.length();++i) a[i] = s1[i] - '0';
	for (int i = 0;i < s2.length();++i) b[i] = s2[i] - '0';
	//to turn the two strings into two arrays
	reverse(a,a + s1.length());
	reverse(b,b + s2.length());
	int len = max(s1.length(),s2.length());
	//to get the maximum of the two lengths of the strings
	//that might be the length of the answer
	for (int i = 0;i < len;++i)
	{
		c[i] += a[i] + b[i]; //to add the number of a and b
		c[i + 1] += c[i] / 10; //if a plus b is larger than nine,the next number would plus one
		c[i] %= 10; //after the next number plus one,this one should minus ten
	}
	if (c[len]) ++len;
	//if the last two numbers' sum is bigger than 10,a new number should be developed
	string str = ""; //the string is to record the answer
	for (int i = 0;i < len;++i)
	{
		char ch = c[i] + 48; //to turn the number into a char
		str = ch + str; //let the string plus the char
	}
	return str; //return the value of the string as the answer
}

int main()
{
	string s1,s2;
	getline(cin,s1);
	getline(cin,s2);
	//to read the two numbers as strings,you can also use cin to do the work
	s1 = add(s1,s2); //to cover the value of s1 with the answer
	cout << s1 << endl; //to output the answer with cout,you can also use printf to do the work
}
